org.apache.jasper.JasperException: For input string: "1</a>_J2SE

jsp 代码错误
type Exception report

message

description The server encountered an internal error () that prevented it from fulfilling this request.

exception

org.apache.jasper.JasperException: For input string: "1</a>"
org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:372)
org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:292)
org.apache.jasper.servlet.JspServlet.service(JspServlet.java:236)
javax.servlet.http.HttpServlet.service(HttpServlet.java:802)

root cause

java.lang.NumberFormatException: For input string: "1</a>"
java.lang.NumberFormatException.forInputString(NumberFormatException.java:48)
java.lang.Integer.parseInt(Integer.java:456)
java.lang.Integer.parseInt(Integer.java:497)
org.apache.jsp.reply_jsp._jspService(reply_jsp.java:48)
org.apache.jasper.runtime.HttpJspBase.service(HttpJspBase.java:94)
javax.servlet.http.HttpServlet.service(HttpServlet.java:802)
org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:324)
org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:292)
org.apache.jasper.servlet.JspServlet.service(JspServlet.java:236)
javax.servlet.http.HttpServlet.service(HttpServlet.java:802)

note The full stack trace of the root cause is available in the Apache Tomcat/5.0.28 logs.

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Apache Tomcat/5.0.28

请问一下这是什么错误啊，代码那么长不知道改哪里啊

------解决方案--------------------
.NumberFormatException

你想转化为整数，但是无法转为整数，因为有非法符号
------解决方案--------------------
java.lang.NumberFormatException: For input string: "1</a>"
java.lang.NumberFormatException.forInputString(NumberFormatException.java:48)

这错误很明显吧、

org.apache.jsp.reply_jsp._jspService(reply_jsp.java:48)
把48行贴出来看看。
------解决方案--------------------
java.lang.NumberFormatException: For input string: "1</a>"
java.lang.NumberFormatException.forInputString(NumberFormatException.java:48)
根据这两句可以看出数据类型转换有问题！格式不对，或者出现了空值
------解决方案--------------------

探讨

能不能具体到那个语句啊，我看着都对

------解决方案--------------------
out.println(request.getParameter("id"));
out.println(request.getParameter("rootId"));

先看看传过来的参数是什么值，先判断 null 和"" 嘛

------解决方案--------------------

Java code

//从上个页面传过来的id或者rootId含有非数字格式的数据，多了个</a>，按楼上说的先把这俩参数打印出来看看<% int id = Integer.parseInt(request.getParameter("id"));int rootId = Integer.parseInt(request.getParameter("rootId")); %>
------解决方案--------------------
探讨

就是多了个</a>知道啦。谢啦

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